JEE Main 2022 — Mathematics Calculus

Given,

$$(1+{e}^{2x})(\frac{dy}{dx}+y)=1$$(1+e2x)(dxdy​+y)=1

Now on rearranging we get,

$$\Rightarrow \frac{dy}{dx}+y=\frac{1}{1+{e}^{2x}}$$⇒dxdy​+y=1+e2x1​

We can see it is a linear differential equation,

So integrating factor is $\displaystyle {e}^{\int 1\cdot dx}={e}^{x}$e∫1⋅dx=ex

So solution will be

$$y\cdot IF=\int \frac{1}{1+{e}^{2x}}\times IFdx$$y⋅IF=∫1+e2x1​×IFdx

$$\Rightarrow y{e}^{x}=\int \frac{1}{1+{e}^{2x}}\times {e}^{x}dx$$⇒yex=∫1+e2x1​×exdx

$$\Rightarrow y\cdot {e}^{x}={\mathrm{tan}}^{-1}({e}^{x})+c$$⇒y⋅ex=tan−1(ex)+c

Now as curve is passing through $(0,\frac{\pi }{2})$(0,2π​) so

$$\Rightarrow c=\frac{\pi }{4}$$⇒c=4π​

Now calculating the limit $\underset{x\rightarrow \infty }{\mathrm{lim}}(y\cdot {e}^{x})$x→∞lim​(y⋅ex) we get,

$$\Rightarrow \underset{x\rightarrow \infty }{\mathrm{lim}}(y\cdot {e}^{x})=\underset{x\rightarrow \infty }{\mathrm{lim}}({\mathrm{tan}}^{-1}({e}^{x})+\frac{\pi }{4})=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$$⇒x→∞lim​(y⋅ex)=x→∞lim​(tan−1(ex)+4π​)=2π​+4π​=43π​